Question 1180602


The length of a rectangle is {{{5}}} feet longer the net with 

{{{L=W+5}}}...eq.1


if the perimeter of the rectangle it’s {{{50}}}, we have

{{{2(L+W)=50}}}

{{{L+W=25}}}..........substitute {{{L}}} from eq.1

{{{W+5+W=25}}}

{{{2W=25-5}}}

{{{2W=20}}}

{{{W=10ft}}}

go to

{{{L=W+5}}}...eq.1.......substitute {{{W}}}

{{{L=10+5}}}

{{{L=15ft}}}


then,  the area is

{{{A=LW}}}

{{{A=15ft*10ft}}}

{{{A=150ft^2}}}